Question

Into a circular drum of radius 4.9m and height 3.6m, how many full bags of wheat can be emptied if the space required for wheat in each bag is $1.8{m^3}?$ (take $\pi  = 3.14$)

Answer 1

Hint: First we have to define what the terms we need to solve the problem are.

Since a circular drum is given as the radius of $4.9m$ which is half the diameter $d = \dfrac{r}{2}$ (so the total diameter of the drum is $9.8m$) and also the height is given as $3.6m$ for the circular drum, pie value can be written as $\pi  = 3.14$.

Formula used: $\pi {(r)^2}h$ (volume of the drum)

Complete step-by-step solution:

Since the volume of the drum is $\pi {(r)^2}h$; now substitute the known values into the formula we get $\pi {(r)^2}h \Rightarrow 3.14 \times {(4.9)^2} \times 3.6$ 

Since the volume of each bag of the wheat is given as $1.8{m^3}$; to find the number of bags that we need to fit into that space required of the circular drum can be found by dividing the volume of the drum to the volume of bags;

Hence the number of bags $ \Rightarrow \dfrac{{3.14 \times {{(4.9)}^2} \times 3.6}}{{1.8}}$ 

Thus, further solving we get the number of bags $ \Rightarrow \dfrac{{3.14 \times 4.9 \times 4.9 \times 3.6}}{{1.8}} = \dfrac{{271.4}}{{1.8}}$

Hence the number of bags that required is $ \Rightarrow \dfrac{{271.4}}{{1.8}} = 150.7$(meter cube) approximately $151$ bags that we will need to fill that circular drum.

Note:  Since the height and radius of the problem are given so that the required answer is the number of bags, if the number of bags and height is given, we can apply the same formula and methods to find the radius of the problem. Since $\pi  = 3.14$ is an irrational number because it was not in the $\dfrac{p}{q}$ format.